PUMaC 2020 · 个人决赛（B 组） · 第 3 题
PUMaC 2020 — Individual Finals (Division B) — Problem 3
题目详情
- Let ABC be a triangle and let the points D, E be on the rays AB, AC such that BCED is cyclic. Prove that the following two statements are equivalent: • There is a point X on the circumcircle of ABC such that BDX , CEX are tangent to each other. 2 • AB · AD ≤ 4 R , where R is the radius of the circumcircle of ABC . 1
解析
- Let ABC be a triangle and let the points D, E be on the rays AB, AC such that BCED is cyclic. Prove that the following two statements are equivalent: • There is a point X on the circumcircle of ABC such that BDX , CEX are tangent to each other. 2 • AB · AD ≤ 4 R , where R is the radius of the circumcircle of ABC . Solution : Let X be an arbitrary point on the circumcircle of ABC , and let Y be the intersection of circles BDX and CEX different from X (if the two circles are tangent, set Y = X ). Then, when X → B , then Y → D , and similarly X → C then Y → E . Moreover, as X continuously goes over the arc BC , Y must move continuously as well. Now, we have a lemma: Lemma: Y ∈ DE . Proof. This follows by angle chasing or by Miquel’s thm applied on the triangle ADE and circles ABC, BDX, CEX . Here is the angle-chasing solution (we assume oriented angles): ◦ ◦ ◦ ∠ DY E = ∠ DY X + ∠ XY E = 180 − ∠ XBD + 180 − ∠ ECX = ∠ ABX + ∠ XCA = 180 . The previous two facts thus give that Y traces the segment DE as X moves from B to C (and possibly something more). Now, we have the claim that completes the proof: There is a point X that satisfies 1) iff DE intersects the circumcircle Γ of ABC . Namely, if DE intersects Γ at Y , pick X that gives 0 Y = Y . Then, we must have X = Y , or X = B , as otherwise BXD intersects ABC in three 0 points. X = B is not possible, so we must have X = Y , and BDX is tangent to CEX . In the other direction, if there is a point X satisfying 1), we must have X = Y ∈ DE , and then X ∈ Γ ∩ DE . 2 ′ The final step is to prove that DE intersects ABC iff AB · AD ≤ 4 R . Let A be the antipode ′ of A on Γ. It is clear that the tangent to Γ at A is parallel to DE . Thus, we have that DE ′ ′ ′ meets Γ iff DE ∩ AA lies on the segment AA . Finally, this holds iff D is on the segment AD , ′ ′ there D is the foot of the perpendicular from A onto AB , and this happens precisely when ′ ′ 2 2 AD · AB ≤ AD · AB = A A = 4 R . Proposed by Aleksa Milojevi´ c. 2